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How many cases of a three digit number $ABC$ exist such that the product of the two two-digit numbers $AB$ and $BC =$ the three digit number of $ABC$. $A, B,$ and $C$ represent the digits of the numbers in their respective places.

So far I've came up with the following equation:

$$100a + 10b + c = (10a + b)(10b + c),$$

but that would give you a messy equation with multiple variables that wouldn't really factor or simplify.

How else would you find out the number of possibilities of $ABC$ using only logic and reasoning?

Thanks.

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Note that if $AB \times BC = ABC$, then since $AB$ is a multiple of $AB0$, it follows that $AB$ divides the difference $ABC - AB0 = C$. But since $AB > C$, we find this is impossible...

EDIT : Unless $C = 0$ (thanks to a comment below). In which case, obviously we get $B = 1$, and $A $ can be of our choice (example : $31 \times 10 = 310$ etc.) These are clearly all the possibilities.

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    $\begingroup$ This is not impossible. Take for example, $210=21*10$. $\endgroup$ – greenturtle3141 Oct 14 '17 at 3:08
  • $\begingroup$ Yes, thank you for pointing this out, I have cleared it up. $\endgroup$ – астон вілла олоф мэллбэрг Oct 14 '17 at 3:11
  • $\begingroup$ Thank you. I now understand why C = 0. But why must that necessitate B = 1? $\endgroup$ – user491229 Oct 14 '17 at 4:05
  • $\begingroup$ If $C = 0$, then the solution becomes $ab \times b0 = ab0$. Now, we can expand these out, to get $(10a+b)(10b) = 100a + 10b$, which simplifies to $100ab + 10b^2 = 100a + 10b$. Suppose $b > 1$, then note that $ab > a$ and $b^2 > b$, so we should expect that $100ab > 100a$ and $10b^2 > 10b$. However, summing these, we get $100ab+ 10b^2 > 100a + 10b$, which is a contradiction. Hence, $b=0,1$ follows. But $b=0$ gives the left side as zero, which is smaller than the right side as $a > 0$ is necessary for $abc$to be a three digit number. Hence, $b=1$. $\endgroup$ – астон вілла олоф мэллбэрг Oct 14 '17 at 4:12

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